Problem: Simplify and expand the following expression: $ \dfrac{4}{4k - 28}- \dfrac{5}{4k - 12}- \dfrac{k}{k^2 - 10k + 21} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{4}{4k - 28} = \dfrac{4}{4(k - 7)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4k - 12} = \dfrac{5}{4(k - 3)}$ We can factor the quadratic in the third term: $ \dfrac{k}{k^2 - 10k + 21} = \dfrac{k}{(k - 7)(k - 3)}$ Now we have: $ \dfrac{4}{4(k - 7)}- \dfrac{5}{4(k - 3)}- \dfrac{k}{(k - 7)(k - 3)} $ The least common multiple of the denominators is: $ 16(k - 7)(k - 3)$ In order to get the first term over $16(k - 7)(k - 3)$ , multiply by $\dfrac{4(k - 3)}{4(k - 3)}$ $ \dfrac{4}{4(k - 7)} \times \dfrac{4(k - 3)}{4(k - 3)} = \dfrac{16(k - 3)}{16(k - 7)(k - 3)} $ In order to get the second term over $16(k - 7)(k - 3)$ , multiply by $\dfrac{4(k - 7)}{4(k - 7)}$ $ \dfrac{5}{4(k - 3)} \times \dfrac{4(k - 7)}{4(k - 7)} = \dfrac{20(k - 7)}{16(k - 7)(k - 3)} $ In order to get the third term over $16(k - 7)(k - 3)$ , multiply by $\dfrac{16}{16}$ $ \dfrac{k}{(k - 7)(k - 3)} \times \dfrac{16}{16} = \dfrac{16k}{16(k - 7)(k - 3)} $ Now we have: $ \dfrac{16(k - 3)}{16(k - 7)(k - 3)} - \dfrac{20(k - 7)}{16(k - 7)(k - 3)} - \dfrac{16k}{16(k - 7)(k - 3)} $ $ = \dfrac{ 16(k - 3) - 20(k - 7) - 16k} {16(k - 7)(k - 3)} $ Expand: $ = \dfrac{16k - 48 - 20k + 140 - 16k}{16k^2 - 160k + 336} $ $ = \dfrac{-20k + 92}{16k^2 - 160k + 336}$ Simplify: $ = \dfrac{-5k + 23}{4k^2 - 40k + 84}$